Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 12 - Kinematics of a Particle - Section 12.2 - Rectilinear Kinematics: Continuous Motion - Problems - Page 19: 23

Answer

$s=10(1-e^{-2t})$ m $v=(20e^{-2t})$ m/s $a=\frac{dv}{dt}=(-40e^{-2t})$ m/s$^2$

Work Step by Step

$a=-2v$ $\frac{dv}{dt}=-2v$ $\int_{20}^v \frac{dv}{v}=\int_0^t -2 dt$ $\ln \frac{v}{20}=-2 t$ $v=(20e^{-2t})$ m/s $a=\frac{dv}{dt}=(-40e^{-2t})$ m/s$^2$ $\int_0^s ds = v dt = \int_0^t(20e^{-2t})dt$ $s=-10e^{-2t}|^t_0=-10(e^{-2t}-1)$ $s=10(1-e^{-2t})$ m
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